– WELCOME TO A LESSON

ON SOLVING FOR TIME IN THE ANNUITY AND LOAN FORMULAS

USING COMMON LOGARITHMS. THIS LESSON DOES ASSUME

YOU’VE ALREADY WATCHED THE VIDEO ON HOW TO SOLVE EXPONENTIAL

EQUATIONS USING COMMON LOGARITHMS. OFTEN WE ARE INTERESTED IN HOW LONG IT WILL TAKE

TO ACCUMULATE MONEY OR HOW LONG WE NEED

TO EXTEND A LOAN TO BRING PAYMENTS DOWN

TO A REASONABLE LEVEL. THIS REQUIRES SOLVING FOR TIME

IN THE ANNUITY OR LOAN FORMULAS, WHICH WE SEE HERE BELOW. IN BOTH CASES,

CAPITAL N IS TIME IN YEARS. TO SOLVE FOR N,

WE’LL USE LOGARITHMS. MORE SPECIFICALLY, WE’LL BE

USING COMMON LOGARITHMS. IN THIS LESSON WE’LL BE USING

THE ANNUITY FORMULA, AND THEN WE’LL SHOW

HOW TO SET UP A SIMILAR PROBLEM USING THE LOAN FORMULA. LET’S BEGIN BY REVIEWING

WHAT THE VARIABLES REPRESENT IN THE ANNUITY FORMULA. P SUB N IS THE BALANCE

IN THE ACCOUNT AFTER N YEARS. D IS THE REGULAR DEPOSIT

OR DEPOSIT EVERY TIME PERIOD. R IS THE ANNUAL INTEREST RATE

AS A DECIMAL, AND K IS THE NUMBER OF

COMPOUNDING PERIODS IN ONE YEAR. LET’S TAKE A LOOK

AT OUR EXAMPLE. IF YOU INVEST $200 PER MONTH

IN AN ACCOUNT THAT PAYS 3.5% COMPOUNDED

MONTHLY INTEREST, HOW LONG WILL IT TAKE FOR THE

ACCOUNT TO GROW TO $12,000? LET’S BEGIN BY LISTING OUT

THE GIVEN INFORMATION. IF YOU’RE INVESTING $200 PER

MONTH, THAT MEANS THAT D IS 200. THE INTEREST RATE IS 3.5%. AS A DECIMAL THAT WOULD BE

0.035, WHICH IS R. INTEREST IS COMPOUNDED MONTHLY, AND SINCE THERE ARE 12 MONTHS

IN A YEAR, K IS 12. WE WANT TO KNOW HOW LONG

IT WILL TAKE FOR THE ACCOUNT TO GROW TO $12,000. SO P SUB N WOULD BE 12,000. WE DON’T KNOW THE TIME. SO WE’RE TRYING TO SOLVE FOR N, AND NOW, WE’LL SUB THESE VALUES

INTO OUR EQUATION AND THEN SOLVE FOR N. SO NOTICE THAT D=200, K=12, WHICH OCCURS HERE, HERE

AND HERE. R=0.035,

WHICH IS HERE AND HERE, AND P SUB N=12,000. SO AGAIN, OUR GOAL HERE IS TO

SOLVE THIS FOR N, AND BECAUSE N

IS IN THE EXPONENT, THIS IS WHY WE NEED LOGARITHMS

TO SOLVE THIS EQUATION. NOW, TO BEGIN, NOTICE

HOW WE HAVE THIS FRACTION HERE IN THE DENOMINATOR, AND THIS FRACTION BAR MEANS

DIVISION. SO NOTICE HOW IF WE MULTIPLY

BOTH SIDES OF THE EQUATION BY THIS FRACTION HERE, WE CAN ELIMINATE THIS FRACTION

FROM THE RIGHT SIDE. AGAIN, IN THIS FIRST STEP, WE MULTIPLY BOTH SIDES OF THE

EQUATION BY THIS FRACTION HERE. IT’S HELPFUL IF WE CAN THINK OF

THIS AS BEING OVER ONE, AND THEREFORE,

THESE TWO SIMPLIFY TO ONE, AND THIS FRACTION x 12,000=35. LET’S JUST VERIFY THAT. WE HAVE .035 DIVIDED BY 12

AND THEN x 12,000. SO THAT’S WHERE THE 35

CAME FROM. NOW, FOR THE NEXT STEP, IT SAYS

200 x THIS QUANTITY HERE. SO LET’S GO AHEAD AND DIVIDE

BOTH SIDES BY 200, WHICH MEANS THIS WILL SIMPLIFY

TO ONE. SO WE CAN DROP THESE OUTER

PARENTHESES, AND WE’LL LEAVE THIS

AS A FRACTION, EVEN THOUGH WE COULD SIMPLIFY

IT. SO NOW, WE HAVE 35 DIVIDED

BY 200=THIS QUANTITY HERE. AGAIN, NOTICE HOW THIS IS RAISED

TO THE 12N POWER. SO FOR THE NEXT STEP,

WE WILL ADD ONE TO BOTH SIDES. SO WE’D HAVE 35/200 + 1. AGAIN, WE COULD TRY TO SIMPLIFY

THIS, BUT I’M NOT GOING TO, AND THIS IS EQUAL TO THE

QUANTITY 1 + .035 DIVIDED BY 12 RAISED TO THE POWER OF 12N. THIS IS WHERE THE COMMON LOG

IS GOING TO HELP US SOLVE THIS. LET’S GO AHEAD AND PUT THIS

IN PARENTHESES, TOO. WE’RE GOING TO TAKE

THE COMMON LOG OF BOTH SIDES OF THE EQUATION. BY DOING THIS, WE CAN APPLY THE

POWER PROPERTY OF LOGARITHMS HERE, MEANING WE CAN TAKE

THE EXPONENT OF 12N AND MOVE IT TO THE FRONT. SO WE’D HAVE 12N

x THIS LOGARITHM. SO NOW, WE HAVE LOG OF 35

DIVIDED BY 200 + 1=12N x LOG OF 1 + .035

DIVIDED BY 12, AND NOW,

WE CAN ACTUALLY SOLVE FOR N. WE WANT TO ISOLATE N. WE DON’T WANT THIS 12 HERE

OR THIS LOG, WHICH MEANS WE CAN NOW DIVIDE

BOTH SIDES BY 12 LOG OF THE QUANTITY 1

+ .035 DIVIDED BY 12, AND THIS WILL GIVE US

THE VALUE OF N. NOTICE IF WE SIMPLIFY THE RIGHT

SIDE, 12/12 SIMPLIFIES TO 1, AND THESE TWO LOGS ARE THE SAME. SO THIS ALSO SIMPLIFIES TO ONE. WE COULD HAVE CONVERTED SOME OF

THIS TO DECIMALS EARLIER, BUT THEN IF WE USE

ROUNDED DECIMALS TO PERFORM MORE CALCULATIONS, OUR ANSWER WOULD NOT BE

AS ACCURATE. SO NOW, ON THE RIGHT SIDE

WE JUST HAVE N, AND TO EVALUATE THE LEFT SIDE,

WE’LL GO BACK TO THE CALCULATOR. WE DO WANT TO PUT THE ENTIRE

NUMERATOR AND DENOMINATOR IN ITS OWN SET OF PARENTHESES. SO WE’LL HAVE AN OPEN

PARENTHESIS COMMON LOG, WHICH IS HERE, AND THEN WE HAVE

35 DIVIDED BY 200 + 1. NOW, WE’LL NEED A CLOSE

PARENTHESIS FOR THE LOGARITHM AND ANOTHER CLOSE PARENTHESIS

FOR THE NUMERATOR AND THEN DIVIDED BY OPEN

PARENTHESIS FOR THE DENOMINATOR, 12 COMMON LOG 1 + .035

DIVIDED BY 12, CLOSE PARENTHESIS

FOR THE LOGARITHM, ANOTHER CLOSE PARENTHESIS

FOR THE DENOMINATOR AND ENTER. SO IF WE ROUND TO THE NEAREST

TENTH, WE CAN SAY THIS WOULD TAKE

APPROXIMATELY 4.6 YEARS. SO ONCE AGAIN, THIS MEANS THAT

IF YOU INVEST $200 PER MONTH IN AN ACCOUNT THAT PAYS 3.5%

COMPOUNDED MONTHLY INTEREST, IN APPROXIMATELY 4.6 YEARS, THE BALANCE WOULD BE

APPROXIMATELY $12,000. NOW, THIS EXAMPLE

WAS FOR AN ANNUITY, BUT THE PROCESS WOULD BE THE

SAME IF WE USE A LOAN FORMULA, EVEN THOUGH THE EQUATION WOULD

LOOK A LITTLE BIT DIFFERENT. SO BEFORE WE GO, LET’S AT LEAST SET ONE UP

THAT INVOLVES A LOAN FORMULA. FOR EXAMPLE, IF YOU PURCHASE A

$799 TABLET ON YOUR CREDIT CARD, WHICH HAS AN INTEREST RATE

OF 14% COMPOUNDED MONTHLY, HOW LONG WILL IT TAKE FOR YOU

PAY OFF THE PURCHASE IF YOU MAKE PAYMENTS

OF $75 A MONTH? SO USING OUR LOAN FORMULA

THIS TIME, NOTICE THAT P SUB 0 OR

THE PRINCIPAL WOULD BE $799. THE MONTHLY PAYMENT, D, IS $75. THE INTEREST RATE AS A DECIMAL

WOULD BE 0.14, AND BECAUSE THE INTEREST

IS COMPOUNDED MONTHLY, AND THERE’S 12 MONTHS IN A YEAR,

K IS 12. NOW, WE COULD FOLLOW THE SAME

PROCEDURE THAT WE DID FOR THE ANNUITY FORMULA AND SOLVE THIS FOR N

TO ANSWER THE QUESTION. I HOPE YOU FOUND THIS HELPFUL.

but this video doesnt show how to solve for N in a loan formula.